Equipartition of energy in Rayleigh-Jeans law

[vabite]

Goodafternoon everyone,

I am looking to the Rayleigh-Jeans law derivation. In order to calculare the average energy of the independent modes of the EM field in the cavity the equipartition of energy is used. In this way, the average energy of a single mode is found to be <E>=kT.

I ask you which are the two degrees of freedom of an independent mode of an EM field, and (equivalently, I suppose) why such modes can be viewed as harmonic oscillators.

 

[WannabeNewton]

The Fourier modes of the EM field in the cavity have two independent polarization states.

The harmonic oscillator interpretation just comes from the Fourier decomposition of the EM field which is itself just a consequence of it satisfying the free wave equation inside the cavity.

 

[Jano L.]

Goodafternoon everyone,

I am looking to the Rayleigh-Jeans law derivation. In order to calculare the average energy of the independent modes of the EM field in the cavity the equipartition of energy is used. In this way, the average energy of a single mode is found to be <E>=kT.

I ask you which are the two degrees of freedom of an independent mode of an EM field, and (equivalently, I suppose) why such modes can be viewed as harmonic oscillators.

The idea that there are modes and each one is an abstract harmonic oscillator follows from Maxwell's equations for vacuum and boundary conditions for metallic cavity (electric field is zero at the walls) or periodic boundary conditions for cuboid (just a math trick).

EM field can be expressed as a sum of field due to modes, where each mode has variable that says how strong that mode oscillates - amplitude.

Based on the Poynting formula for EM energy, each such mode is ascribed energy in the most natural way. Formally it turns out that the expression is similar to the expression for energy of ordinary harmonic oscillator - it is given by two terms : one proportional to square of the mode amplitude and the other one to square of its time derivative.

And now equipartition is used - each quadratic degree of freedom gets $k_BT/2$ of energy.

 

[Jano L.]

The Fourier modes of the EM field in the cavity have two independent polarization states...

If the polarization is different, it is a different mode :-) Two degrees of freedom per one mode are due to the fact that the energy has two quadratic contributions.

 

[vabite]

It makes sense, thank you.
Can you just write me down the formula for the energy of a mode of the EM field, containing the two quadratic contributions?

 

[Serianox]

The idea that there are modes and each one is an abstract harmonic oscillator follows from Maxwell's equations for vacuum and boundary conditions for metallic cavity (electric field is zero at the walls) or periodic boundary conditions for cuboid (just a math trick).

EM field can be expressed as a sum of field due to modes, where each mode has variable that says how strong that mode oscillates - amplitude.

Based on the Poynting formula for EM energy, each such mode is ascribed energy in the most natural way. Formally it turns out that the expression is similar to the expression for energy of ordinary harmonic oscillator - it is given by two terms : one proportional to square of the mode amplitude and the other one to square of its time derivative.

And now equipartition is used - each quadratic degree of freedom gets $k_BT/2$ of energy.

I'm struggling with this right now. Could you please write the formula for the energy of each Fourier mode where both quadratic terms Appear? Where does the time derivative come from?
I'm losing my mind with all these poor explanations from the books saying that the modes are harmonic oscillators. Thanks a lot